How to compute Ext over an exterior algebra I found this question in several places (even on mathoverflow and mathstackexchange), but I never found a satisfying answer. 

Let $k$ be a field and $V$ a finite dimensional $k$-vectorspace. I would like to compute $\mathrm{Ext}_{\bigwedge V}(k,k)$. 

By looking online it seems that it should be the dual of the symmetric algebra. The first step it's computing a free resolution of $k$ over $\bigwedge V$. Always by looking online it seems that this resolution should look like
$$K_.:\cdots\rightarrow S^{k+1}V\otimes_k\bigwedge V\xrightarrow{d^{k+1}}S^kV\otimes_k\bigwedge V\rightarrow\cdots,$$
where by $SV$ I'm denoting the symmetric algebra. I think the differential should be 
$$d^{k+1}(v_1\cdots v_{k+1}\otimes w)=\sum_i (-1)^{i+1}v_1\cdots\hat{v}_i\cdots v_{k+1}\otimes v_i\wedge w,$$
but I'm not entirely sure. In fact, it is not obvious to me that this complex is acyclic. Could you tell me how to prove it?
And once we know that this is a resolution how do we compute the Ext's? It seems that the differentials of $\mathrm{Hom}_{\bigwedge V}(K_.,k)$ are zero but I can't understand why. Any help, please?
(I'm also using commutative-algebra and modules as tags since this question is related to the Koszul complex.)
 A: I am far from an expert on Ext, but I can explain why the Koszul complex is acyclic.
The subspaces $S^1 V$ of $SV$ and $\Lambda^1 V$ of $\Lambda V$ are canonically isomorphic. Let $d\colon \Lambda^1 V \to S^1 V$ be such an isomorphism, and $h$ its inverse.
Then both $d$ and $h$ extend uniquely to antiderivations on $KV := SV \otimes \Lambda V$,
of respective degrees $1$ and $-1$,
where we set $\deg \Lambda^1 V = 1$ and $\deg S^1 V = 2$. (Since you want $d$ to be a differential, $d S^1 V = d^2 \Lambda^1 V = 0$, and then it's defined on all the algebra generators, so the product rule does the rest.)
We can also "multiplicatively" grade $KV$ by letting $K^n V$ be spanned by products of $n$ elements of $V$ (in either tensor-factor), so the sum of the subspaces $S^j V \otimes \Lambda^{n-j} V$. Note that on $K^1 V$, we have
$$hd+dh = \mathrm{id}.$$
It's not hard to show inductively that on $K^n V$,
$$hd+dh = n\cdot \mathrm{id}.$$
These chain homotopies then witness the acyclicity of both the DGAs $(KV,d)$ and $(KV,h)$.
